機器學習作業（一）線性回歸——Python(numpy)實現

題目太長啦！文檔下載【傳送門】

第1題

簡述：設計一個5*5的單位矩陣。

import numpy as np
A = np.eye(5)
print(A)

 

運行結果：

 

第2題

簡述：實現單變量線性回歸。

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

#-----------------計算代價值函數-----------------------
def computeCost(X, y, theta):
    m = np.size(X[:,0])
    J = 1/(2*m)*np.sum((np.dot(X,theta)-y)**2)
    return J


#----------------根據人口預測利潤----------------------
#讀取數據集中數據，第一列是人口數據，第二列是利潤數據
data = np.loadtxt('ex1data1.txt',delimiter=",",dtype="float")
m = np.size(data[:,0])
# print(data)

#------------------繪制樣本點--------------------------
X = data[:,0:1]
y = data[:,1:2]
plt.plot(X,y,"rx")
plt.xlabel('Population of City in 10,000s')
plt.ylabel('Profit in $10,000s')
# plt.show()

#-----------------梯度下降計算局部最優解----------------
#添加第一列1
one = np.ones(m)
X = np.insert(X,0,values=one,axis=1)
# print(X)

#設置α、迭代次數、θ
theta = np.zeros((2,1))
iterations = 1500
alpha = 0.01

#梯度下降，並顯示線性回歸
J_history = np.zeros((iterations,1))
for iter in range(0,iterations):
    theta = theta - alpha/m*np.dot(X.T,(np.dot(X,theta)-y))
    J_history[iter] = computeCost(X,y,theta)
plt.plot(data[:,0],np.dot(X,theta),'-')
plt.show()
# print(theta)
# print(J_history)

#--------------------顯示三維圖------------------------
theta0 = np.linspace(-10,10,100)
theta1 = np.linspace(-1,4,100)
J_vals = np.zeros((np.size(theta0),np.size(theta1)))
for i in range(0,np.size(theta0)):
    for j in range(0,np.size(theta1)):
        t = np.asarray([theta0[i],theta1[j]]).reshape(2,1)
        J_vals[i,j] = computeCost(X,y,t)
# print(J_vals)
J_vals = J_vals.T    #需要轉置一下，否則軸會反
fig1 = plt.figure()
ax = Axes3D(fig1)
ax.plot_surface(theta0,theta1,J_vals,rstride=1,cstride=1,cmap=plt.get_cmap('rainbow'))
ax.set_xlabel('theta0')
ax.set_ylabel('theta1')
ax.set_zlabel('J')
plt.show()

#--------------------顯示輪廓圖-----------------------
lines = np.logspace(-2,3,20)
plt.contour(theta0,theta1,J_vals,levels = lines)
plt.xlabel('theta0')
plt.ylabel('theta1')
plt.plot(theta[0],theta[1],'rx')
plt.show()

 

運行結果：

 

 

第3題

簡述：實現多元線性回歸。

import numpy as np
import matplotlib.pyplot as plt

#-----------------計算代價值函數-----------------------
def computeCost(X, y, theta):
    m = np.size(X[:,0])
    J = 1/(2*m)*np.sum((np.dot(X,theta)-y)**2)
    return J

#-------------------根據面積和卧室數量預測房價----------
#讀取數據集中數據，第一列是面積數據，第二列是卧室數量，第三列是房價
data = np.loadtxt('ex1data2.txt',delimiter=",",dtype="float")
m = np.size(data[:,0])
# print(data)
X = data[:,0:2]
y = data[:,2:3]

#----------------------均值歸一化---------------------
mu = np.mean(X,0)
sigma = np.std(X,0)
X_norm = np.divide(np.subtract(X,mu),sigma)
one = np.ones(m)   #添加第一列1
X_norm = np.insert(X_norm,0,values=one,axis=1)
# print(mu)
# print(sigma)
# print(X_norm)

#----------------------梯度下降-----------------------
alpha = 0.05
num_iters = 100
theta = np.zeros((3,1));
J_history = np.zeros((num_iters,1))
for iter in range(0,num_iters):
    theta = theta - alpha/m*np.dot(X_norm.T,(np.dot(X_norm,theta)-y))
    J_history[iter] = computeCost(X_norm,y,theta)
# print(theta)
x_col = np.arange(0,num_iters)
plt.plot(x_col,J_history,'-b')
plt.xlabel('Number of iterations')
plt.ylabel('Cost J')
plt.show()

#----------使用上述結果對[1650,3]的數據進行預測--------
test1 = [1,1650,3]
test1[1:3] = np.divide(np.subtract(test1[1:3],mu),sigma)
price = np.dot(test1,theta)
print(price)        #輸出預測結果[292455.63375132]

#-------------使用正規方程法求解----------------------
one = np.ones(m)
X = np.insert(X,0,values=one,axis=1)
theta = np.dot(np.dot(np.linalg.pinv(np.dot(X.T,X)),X.T),y)
# print(theta)
price = np.dot([1,1650,3],theta)
print(price)       #輸出預測結果[293081.46433497]

 

運行結果：【一個疑惑>>兩種方法求解的估算價格很小，但θ相差較大？】